Home page for accesible maths 1 Real Numbers 1.1 The rationals and the decimals 1.3 Real numbers are just the decimals

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1.2 The mysterious sequences of Monsieur Cauchy

Let $x=x_{0}.x_{1}x_{2}\dots$ be a decimal. Then we call the rational number $x_{0}.x_{1}x_{2}\dots x_{k}$ the $k$ -truncation of $x$ and denote it by $(x)_{k}$ . Clearly, the numbers are getting closer and closer to each other: if $k>l$ then $|x_{k}-x_{l}|\leq 10^{-l}$ . The following definition is due to Baron Augustin-Louis Cauchy, son of Louis-Francois Cauchy, who were quite obviously French people. Cauchy wrote the first rigorous book on mathematical analysis, titled “Cours d’Analyse”.

Definition 1.2.1 (Plain English Version)

A sequence of rational numbers $\{q_{n}\}^{\infty}_{n=1}$ is a Cauchy-sequence if they are getting closer and closer to each other.

Again, we need to translate from Plain English to a More Mathematical form. We wish that if both $n$ and $m$ are large, then $q_{n}$ and $q_{m}$ are closer to each other than a certain number. So, similarly to the definition of convergence, we can come up with the following definition.

Definition 1.2.2 (More Mathematical Version)

A sequence of rational
numbers $\{q_{n}\}^{\infty}_{n=1}$ is a Cauchy-sequence if for all $\varepsilon>0$ there exists $N>0$ such that $|q_{n}-q_{m}|\leq\varepsilon$ provided that $n,m\geq N$ .

So in particular, if $x=x_{0}.x_{1}x_{2}x_{3}\dots$ then as we saw above, $\{(x)_{k}\}^{\infty}_{k=1}$ is a Cauchy-sequence. Suppose that two lions are very close to a tiger. What we know about the lions? OK, they might not be as close to each other as to the tiger, but they must be pretty close to each other. This kindergarten style observation leads to the following proposition.

Proposition 1.2.1 (CONVERGENCE IMPLIES CAUCHY)

A sequence $\{q_{n}\}^{\infty}_{n=1}$
of rational numbers that is converging to a rational number $q$ is a Cauchy-sequence.

Proof: We need to “mathematize” the lions vs. tiger idea. Let $\varepsilon>0$ be a rational number. By the definition of convergence, there exists $N>0$ such that if $n\geq N$ then $|q_{n}-q|<\frac{\varepsilon}{2}$ That is a classical trick in math. You need to prove something for $\varepsilon$ and you prepare for it by proving something for $\frac{\varepsilon}{2}$ , in order to have greater leverage. Hence, if $n,m\geq N$ , then $|q_{n}-q|<\frac{\varepsilon}{2}$ and $|q_{m}-q|<\frac{\varepsilon}{2}$ ( $q_{n},q_{m}$ are the lions, and $q$ is the tiger). Remember that $|a-b|\leq|a|+|b|$ for all rational numbers $a$ and $b$ . So, if $a=q_{n}-q$ , $b=q_{m}-q$ , then $|q_{n}-q_{m}|\leq|q_{n}-q|+|q_{m}-q|<\varepsilon.$ $\square$

Proposition 1.2.2 (CAUCHY IMPLIES BOUNDED)

A Cauchy-sequence $\{q_{n}\}^{\infty}_{n=1}$ of rational numbers is bounded, that is there exists some integer $M$ such that for any $n\geq 1$ , $|q_{n}|\leq M$ .

Proof: By definition, there exists $N>1$ such that $|a_{N}-a_{n}|<1$ if $n\geq N$ . Let $K$ be the maximum of the numbers $|a_{1}|,|a_{2}|,\dots,|a_{N}|$ . Then for any $n\geq 1$ , $|a_{n}|\leq K+1$ . $\square$

As we remarked earlier, it is not so easy to add, subtract or multiply decimals. On the other hand, it is very easy to add, subtract or multiply Cauchy-sequences.

Proposition 1.2.3 (Calculations with Cauchy-sequences)

Let $\{p_{n}\}^{\infty}_{n=1}$ and $\{q_{n}\}^{\infty}_{n=1}$ be Cauchy-sequences. Then, $\{p_{n}+q_{n}\}^{\infty}_{n=1}$ , $\{p_{n}-q_{n}\}^{\infty}_{n=1}$ and $\{p_{n}q_{n}\}^{\infty}_{n=1}$ are all Cauchy-sequences.

Proof: By Proposition 1.2.2, there exists an $M>0$ such that $|q_{n}|<M,|p_{n}|<M$ for any $n\geq 1$ . Let $\varepsilon>0$ . Fix an integer $N$ such that if $n,m>N$ , then $|q_{n}-q_{m}|<\frac{\varepsilon}{2M}$ and $|p_{n}-p_{m}|<\varepsilon$ . Then $|p_{n}q_{n}-p_{m}q_{m}|=|p_{n}(q_{n}-q_{m})-p_{m}(q_{m}-q_{n})|\leq 2M\frac{% \varepsilon}{2M}\leq\varepsilon$ . Therefore $\{p_{n}q_{n}\}^{\infty}_{n=1}$ is a Cauchy-sequence (the other parts are easier and will be given as workshop exercises). $\square$

For a decimal $x=x_{0}.x_{1}x_{2}x_{3}\dots$ , $\{(x)_{k}\}^{\infty}_{k=1}$ is a Cauchy-sequence. For the converse, we have the following proposition.

Proposition 1.2.4

Let $\overline{q}=\{q_{n}\}^{\infty}_{n=1}$ be a Cauchy-sequence of rational numbers. Then there exists a decimal $x_{0}.x_{1}x_{2}\dots$ , the “limitvalue” of $\overline{q}$ , $L(\overline{q})$ such that $(q_{n}-r_{n})\to 0$ , where $r_{n}=x_{0}.x_{1}x_{2}\dots x_{n}$ . This “limitvalue” is unique (except if it is a “bad” decimal ,then both forms are taken as limit values).

Proof: We have two cases.

Case 1: For every $k$ , “after a while” all $q_{n}$ ’s start with the same sequence $x_{0}.x_{1}x_{2}\dots x_{k}$ . That is, there exists some number $N_{k}>0$ (the lower index suggests that this number depends on $k$ ) such that if $n,m>N_{k}$ , then the first $k$ digits of $q_{n}$ and $q_{m}$ are the same.

Then we immediately obtain the decimal $x_{0}.x_{1}x_{2}\dots$ we are looking for.

Case 2: For some $k$ , there are two different starting sequences $x_{0}.x_{1}x_{2}\dots x_{k}>y_{0}.y_{1}y_{2}\dots y_{k}$ that we see infinitely often as $n\to\infty$ .

So, we have pairs $\{q_{n_{i}},q_{m_{i}}\}^{\infty}_{i=1}$ such that $|q_{n_{i}}-q_{m_{i}}|\to 0$ , nevertheless $q_{n_{i}}$ starts with $x_{0}.x_{1}x_{2}\dots x_{k}$ and $q_{m_{i}}$ starts with $y_{0}.y_{1}y_{2}\dots y_{k}$ . But this could happen only if $q_{n_{i}}\to x_{0}.x_{1}x_{2}\dots x_{k}\overline{0}$ and $q_{m_{i}}\to y_{0}.y_{1}y_{2}\dots y_{k}\overline{9}$ . This implies that actually, $q_{n}\to x_{0}.x_{1}x_{2}\dots x_{k}\overline{0}$ . Therefore the decimal $x_{0}.x_{1}x_{2}\dots$ (and $y_{0}.y_{1}y_{2}\dots y_{k}\overline{9}$ ) satisfies the condition of our proposition. $\square$

Example 1.2.1

The decimal $x=x_{0}.x_{1}x_{2}x_{3}\dots$ is the limitvalue of the Cauchy-sequence $\{x_{n}\}^{\infty}_{n=1}$ . It is a Case 1. example.

Example 1.2.2

$0.51,0.49,0.501,0.499,0.5001,0.4999.....$ . It is a Case 2. example.

Lemma 1.2.1

The Cauchy-sequences $\overline{q}=\{q_{n}\}^{\infty}_{n=1}$ and $\overline{p}=\{p_{n}\}^{\infty}_{n=1}$ have the same limitvalue if and only if $(p_{n}-q_{n})\to 0$ .

Proof: If $\overline{q}$ and $\overline{p}$ have the same limitvalue then there exists an $r\in\mathbb{R}$ such that both $(p_{n}-(r)_{n})\to 0$ and $(q_{n}-(r)_{n})\to 0$ . Hence, for any $\varepsilon>0$ , there exists an $N$ , such that if $n\geq N$ then $|p_{n}-(r)_{n}|<\frac{\varepsilon}{2}$ and $|q_{n}-(r)_{n}|<\frac{\varepsilon}{2}$ . Hence, if $n\geq N$ , then $|p_{n}-q_{n}|<\varepsilon$ . That is, $(p_{n}-q_{n})\to 0$ . Now suppose that $(p_{n}-(r)_{n})\to 0$ and $(p_{n}-q_{n})\to 0$ . Then, we can see as above that $(q_{n}-(r)_{n})\to 0$ . $\square$

Now, we finally show how to add, subtract and multiply decimals.

Definition 1.2.3

Let $x=x_{0}.x_{1}x_{2}\dots$ and $y=y_{0}.y_{1}y_{2}\dots$ are decimals. Then $x+y$ is defined as the “limitvalue” of the Cauchy-sequence $\{(x)_{n}+(y)_{n}\}^{\infty}_{n=1}$ . Similarly, $x-y$ is defined as the “limitvalue” of the Cauchy-sequence $\{(x)_{n}-(y)_{n}\}^{\infty}_{n=1}$ and $x\cdot y$ is defined as the “limitvalue” of the Cauchy-sequence $\{(x)_{n}\cdot(y)_{n}\}^{\infty}_{n=1}$ .

So, now we can square the $\pi$ . We take the Cauchy-sequence $q_{1}=3^{2}=9$ , $q_{2}=(3.1)^{2}=9.61$ , $q_{3}=(3.14)^{2}=9.8596$ , $q_{4}=(3.141)^{2}=9.865881$ , $q_{5}=(3.1415)^{2}=9.86902225$
$q_{6}=(3.14159)^{2}=9.8695877281$ , $q_{7}=(3.141592)^{2}=9.86960029446$ , $q_{8}=(3.1415926)^{2}=9.86960406437$ ….. By Proposition 4.5.2, the digits will stabilize to a decimal, and this decimal will be the square of $\pi$ .